Function-Maximum and Minimum

Question

Given that f is a real valued non-constant differentiable function such that $f(x)f'(x)\le0$ , for all real $x$, then it follows that:

(A) $f^2(x)$ is increasing function

(B) $f^2(x)$ is decreasing function

(C) $f(x)$ has no critical point

(D) $f(x)$ does not have any real root

My approach is as follow

Case 1: $f(x)>0$ and $f(x)$ is monotonically decreasing function

Case 2: $f(x)<0$ and $f(x)$ is monotonically increasing function

In both cases $f(x)$ has imaginary roots and no critical points, hence C and D are correct answer

Lets take the case $F(x)=f^2(x)$ Then $F'(x)=2f(x)f'(x)\le0$

Hence $F(x)$ is monotonically decreasing function. Hence (B) is also correct.

I managed to solve this problem using hypothetical cases , is there any other approach to solve this type of problem.

Answer 1

Since $(f^2)'=2ff'\leqslant0$, $f^2$ is decreasing.

Answer 2

Just because you could construct some classes of functions that fulfill your conditions, and those classes have no real root/critical point, does not mean that there do no exist other functions that do have those properties.

For example

$$f(x)= \begin{cases} x^2 & \text{, if } x \le 0 \\ 0 & \text{, if } x > 0 \\ \end{cases} $$

fulfills the conditions of your problem (if "non-constant" means it takes at least 2 different values), but has many real roots, so (D) is wrong (it does not follow from the problem conditions).

(C) is similiarly wrong, as

$$f'(x)= \begin{cases} 2x & \text{, if } x \le 0 \\ 0 & \text{, if } x > 0 \\ \end{cases} $$

also has lots of roots.